Question

the sum of a two digit number and the number obtained by reversing the order of the digits is 99 find the number if the digits differ by 3

Answer 1

let, the digits are x and y
x-y=3-------(1)
again
(10x+y)+(10y+x)=99
=>10x+y+10y+x=99
=>11x+11y=99
=>x+y=9------(2)
(1)+(2)===>
2x=12
=>x=6
putting the value of x into eqn (2)
6+y=9
=>y=3
so the digits are 6 and 3
hope you got it :)